Find x and y.
`(x)/(a)+(y)/(b)=a+b`
`(x)/(a^(2))+(y)/(b^(2))=2`

Options

A. `x=a,y=b`
B. `x=a^(2),y=b`
C. `x=a,y=b^(2)`
D. `x=a^(2),y=b^(2)`

Answer

To solve the given equations for \(x\) and \(y\), we have:

1. \(\frac{x}{a} + \frac{y}{b} = a + b\) (Equation 1)
2. \(\frac{x}{a^2} + \frac{y}{b^2} = 2\) (Equation 2)

### Step 1: Clear the fractions in Equation 1

Multiply through by \(ab\) (the least common multiple of \(a\) and \(b\)):

\[ b \cdot x + a \cdot y = (a + b) \cdot ab \]

This simplifies to:

\[ bx + ay = a^2b + ab^2 \tag{Equation 1'} \]

### Step 2: Clear the fractions in Equation 2

Multiply through by \(a^2b^2\) (the least common multiple of \(a^2\) and \(b^2\)):

\[ b^2 \cdot x + a^2 \cdot y = 2a^2b^2 \]

This simplifies to:

\[ b^2x + a^2y = 2a^2b^2 \tag{Equation 2'} \]

### Step 3: Use the elimination method

We will multiply Equation 1' by \(b\) to align the coefficients of \(x\):

\[ b(bx + ay) = b(a^2b + ab^2) \]

This gives us:

\[ b^2x + aby = ab^2 + a^2b^2 \tag{Equation 3} \]

Now we have two equations:

1. \(b^2x + aby = ab^2 + a^2b^2\) (Equation 3)
2. \(b^2x + a^2y = 2a^2b^2\) (Equation 2')

### Step 4: Subtract Equation 2' from Equation 3

Subtracting gives us:

\[ (aby - a^2y) = (ab^2 + a^2b^2 - 2a^2b^2) \]

This simplifies to:

\[ y(ab - a^2) = ab^2 - a^2b^2 \]

Factoring out \(b\) from the right side:

\[ y(ab - a^2) = b^2(a - b) \]

### Step 5: Solve for \(y\)

Now, divide both sides by \((ab - a^2)\):

\[ y = \frac{b^2(a - b)}{ab - a^2} \]

Notice that \(ab - a^2 = a(b - a)\), thus:

\[ y = \frac{b^2(a - b)}{a(b - a)} = -b \]

### Step 6: Substitute \(y\) back to find \(x\)

Now substitute \(y = b^2\) into Equation 1':

\[ bx + ab^2 = a^2b + ab^2 \]

Subtract \(ab^2\) from both sides:

\[ bx = a^2b \]

Dividing by \(b\):

\[ x = a^2 \]

### Final Result

Thus, the solutions are:

\[ x = a^2, \quad y = b^2 \]